Change of Basis

[@ゾカリクゾ]

For the curious, the w's are the Lagrange polynomials: en.wikipedia.org/wiki/Lagrange_polynomial#A_perspective_from_linear_algebra

[@weizong8814]

This recitation seems to be a glimpse of a more advanced linear algebra course. It contains a lot of knowledge that we do not know yet.

[@qbtc]

You are right. I've found these recitation problems to be usually more complicated and difficult then the assigned homework on the mit/ocw page.

[@jeevan288]

nope . i understood it based on concepts from previous lectures.

[@SEBASTIAN-ul2uf]

Exactly 😂

[@ashadds]

@@jeevan288nost probably you understood nothing

[@tetsuroda]

7:33 I do not understand why the matrix changes bases as required. The reason why I get confused is probably because w_{i} is not an element but a vector. Can somebody make it clearer?

[@berrycoolcat]

I agree, the notation seems a bit confusing. I think about it like this, if you calculate Ac, where c is the vector holding the coords corresponding to the old basis (1,x,x^2): * When c = (1,0,0), which corresponds to the polynomial 1, Ac = (1,1,1), which agrees with the equation "1 = w1 + w2 + w3" above * When c = (0,1,0), which corresponds to the polynomial x, Ac = (-1,0,1), again agreeing with the equation "x = -w1 + 0 * w2 + w3" * Similarly for c = (0,0,1) You can see that the matrix successfully brings 3 linearly independent vectors in the old basis to the new basis, so it must be the correct "change of basis" matrix

[@NisargJain]

6:32 how did she conclude that? can somebody elaborate that? Thanks.

[@pydro432343]

'1' is a polynom of degree

[@NicolasSchmidMusic]

@@pydro432343 Where can you deduce that a = 1 ?

[@gustavotolson7307]

It comes from the table in kzbin.info/www/bejne/qH7ZY6V6nrSHqas where the two bases (1, x, x²) and (w_1, w_2, w_3) are defined using 3 values of x. You simply put the values of x in the grinder and the values are there.

[@mauriciobarda]

according to the table when x= -1 , 1=1 and w1=1, w2=w3=0 ; when x=0 then 1=1 and w1=w3=0 and w2=1, and when x=1 , 1=1, w1=w2=0 and w3=1 . You conclude that 1=w1+w2+w3 . Same way for the other cases.

[@nerophon]

I found the phrasing of the problem confusing. To me, the table implies that, for any polynomial in the space, alpha = 1 when x = -1. But this is impossible; hence the confusion. I was also surprised by the choice of the differentiation matrix D later in the explanation; the natural choice for me would be the transpose of what was written, so that Dx = d/dx where x and d/dx are column vectors.

[@sedahmo5601]

Nicholas Bellerophon Agree. She should put y(x) and y(-1), y(0), y(1) in the corresponding column of that table instead.

[@krischalkhanal2842]

I think understanding this is the real deal. I tried by my own brute force way, and it did work, but she is using shortcuts here, which I am having hard time wrapping my head around. Much struggle needed.

[@BluesbreakerWyatt]

If you find this difficult, it helps a lot to read related chapters in the book "Introduction to Linear algebra", it cover much more details.

[@maths2bsquared172]

can you mention the authors name,please

[@BluesbreakerWyatt]

@@maths2bsquared172 "Introduction to Linear Algebra" by Gilbert Strang, it is the book by lecturer and the main support book of the course.

[@MikhailSamin]

This is absolutely cursed, I struggled for like half an hour to understand the problem and why is the polynomial at some point is some specific other polynomial, like, ??? The most cursed problem in all the recitations

[@yangpiao3071]

I think the key to understand the whole thing is that w1 w2 w3 form a basis of polynomial, so wi must be a combination of 1 x and x squared.

[@quirkyquester]

what's happening at 6:36 my lorrrdd, please save meeee oops

[@justpaulo]

In part a) she concluded that, for any polynomial p(x), we can find α, β and γ (the coeff in basis w1, w2 & w3) simply by calculating p(-1), p(0) and p(1): α = p(-1) β = p(0) γ = p(1) In particular, you can do the same for the initial basis polynomials 1, x & x². In doing so you are expressing the initial basis in the new w1, w2 & w3 basis. If you're struggling it also helps to find out w1, w2 & w3. I did that and they are given by: w1 = 1/2( x² - 1) w2 = -x² + 1 w3 = 1/2( x² + 1) Note that if you do w1 + w2 + w3 you get 1 indeed. Or if you do 6w1 + 5w2 + 4w3 you get -x+5.

[@Longpan898]

Nice exercise and good job, though maybe a bit fast in the explanation. But eventually, using the various comments, I fall back on my 2 legs. Thanks 🙏🏻

[@dalisabe62]

Not too clear and I am sure the concept should be a fairly easy one explained in more organized steps.

[@kevinshao9148]

oh wow!

[@pydro432343]

entry 2,1 of A inverse is 1/2 (not -1/2) thanks

[@NisargJain]

nope she's right..

[@pydro432343]

@@NisargJain of course she is... (what was I thinking?) thanks

[@kirubelmelak7143]

Entry 3, 2 on the inverse is 0.

[@thedailyepochs338]

i think khan academy does a better job at explaining basis than this recitation. Good lecture though

[@iamjojo999]

Good polynomial examples, but poor explanations. She just go through it, not explain a bit. I can't understand at all until I see comments. She should have written down w1,w2,w3 in polynomial form first. Namely, assume w1=a1×(1)+b1×(x)+ c1× (x^2) And w2=a2×(1)+b2×(x)+ c2× (x^2) And w3=a3×(1)+b2×(x)+ c3× (x^2) So, written above 3 equations into matrix form We have [w1(next row) w2 (next row) w3]= [a1,b1,c1 (next row) a2,b2,c2 (next row) a3,b3,c3 ] × [ 1 (next row) x (next row) x^2] By writting in matrix form, we can easily observe the 3 by 3 matrix of a1 b1 c1 a2 b2 c2 a3 b3 c3 is the matrix that change basis of right hand side (1,x,x^2) to basis of left hand side (w1,w2,w3) So all we need to do is inserting known values of x=-1,0,1 and corresponding w1,w2,w3 at these points into above matrix to get all the coeeficients, which is what she does in part b.